Áp dụng Bunyakovsky, ta có :

\(\left(1+1\right)\left(x^2+y^2\right)\ge\left(x.1+y.1\right)^2=1\)

=> \(\left(x^2+y^2\right)\ge\frac{1}{2}\)

=> \(Min_C=\frac{1}{2}\Leftrightarrow x=y=\frac{1}{2}\)

Mấy cái kia tương tự 

a) \(2x^2-x+1=2\left(x-\dfrac{1}{4}\right)^2+\dfrac{7}{8}\ge\dfrac{7}{8}\)

\(ĐTXR\Leftrightarrow x=\dfrac{1}{4}\)

b) \(5x-x^2+4=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{41}{4}\le\dfrac{41}{4}\)

\(ĐTXR\Leftrightarrow x=\dfrac{5}{2}\)

c) \(x^2+5y^2-2xy+4y+3=\left(x-y\right)^2+\left(2y+1\right)^2+2\ge2\)

\(ĐTXR\Leftrightarrow\)\(x=y=-\dfrac{1}{2}\)

b: ta có: \(-x^2+5x+4\)

\(=-\left(x^2-5x-4\right)\)

\(=-\left(x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{41}{4}\right)\)

\(=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{41}{4}\le\dfrac{41}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{5}{2}\)

a: Ta có: \(A=x^2+3x+4\)

\(=x^2+2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{7}{4}\)

\(=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{3}{2}\)

$A = x(x^2 - 6)$

$A = x^3 - 6x$

Áp dụng bấtt đẳng thức $AM-GM$ ta được:

$x^3 + 2\sqrt2 + 2\sqrt2 \geq 3\sqrt[3]{x^3.8}= 6x$

$\Rightarrow x^3 - 6x \geq - 4\sqrt2$

$\Rightarrow A \geq -4\sqrt2$

Dấu $=$ xảy ra $\Leftrightarrow x^3 = 2\sqrt2 \Leftrightarrow x = \sqrt2$

Vậy $\min A = -4\sqrt2 \Leftrightarrow x =\sqrt2$

đặt A=\(\frac{x^2}{x^2}-\frac{2x}{x^2}+\frac{2013}{x^2}\)\(=\)\(1-2\frac{1}{x}+2013\frac{1}{x^2}\)

đặt \(\frac{1}{x}=a\)\(=>\)\(\frac{1}{x^2}=a^2\)

khi đó \(A=2013a^2-2a+1\)

  \(=>\)\(2013A=\left(2013a\right)^2-4026a+2013\)

                                  \(=\left(2013a-1\right)^2+2012\)

                  bạn tự giải tiếp nhé :))